1

### JEE Main 2018 (Online) 16th April Morning Slot

Assuming ideal gas behaviour, the ratio of density of ammonia to that of hydrogen chloride at same temperature and pressure is : (Atomic wt. of Cl = 35.5 u)
A
1.46
B
0.46
C
1.64
D
0.64

## Explanation

We know, PV = nRT

n = no. of moles = ${m \over M}$

So,   PV = ${m \over M}RT$

$\Rightarrow $$\,\,\,\, P = {m \over V} \times {{RT} \over M} \Rightarrow$$\,\,\,\,$ P = d $\times$ ${{RT} \over M}$ [ d = density = ${m \over V}$ ]

at constant temperature and pressure d $\propto$ M

Now let d1 and d2 are the density of ammonia and HCl.

$\therefore\,\,\,\,$ ${{{d_1}} \over {{d_2}}} = {{{M_{N{H_3}}}} \over {{M_{HCl}}}}$

$\Rightarrow$ $\,\,\,\,$ ${{{d_1}} \over {{d_2}}}$ = ${{17} \over {36.5}}$ = 0.46
2

### JEE Main 2019 (Online) 9th January Morning Slot

0.5 moles of gas A and x moles of gas B exert a pressure of 200 Pa in a container of volume 10 m3 at 1000 K. Given R is the gas constant in JK$-$1mol$-$1, x is :
A
${{2R} \over {4 + R}}$
B
${{2R} \over {4 - R}}$
C
${{4 + R} \over {2R}}$
D
${{4 - R} \over {2R}}$

## Explanation

We know,

PV = nRT

Given,

P = 200 Pa

V = 10 m3

T = 1000 K

n = 0.5 + x

$\therefore$   200 $\times$ 10 = (0.5 + x) R $\times$ 1000

$\Rightarrow$    0.5 + x = ${2 \over R}$

$\Rightarrow$    x = ${2 \over R} - {1 \over 2}$

$\Rightarrow$   x = ${{4 - R} \over {2R}}$
3

### JEE Main 2019 (Online) 12th January Morning Slot

The volume of gas A is twice than that of gas B. The compressibility factor of gas A is thrice than that of gas B at same temperature. The pressures of the gases for equal number of moles are -
A
2PA = 3PB
B
3PA = 2PB
C
PA = 3PB
D
PA = 2PB

## Explanation

Z = PV/nRT

$\Rightarrow$ P = ${{ZnRT} \over V}$

at constant T and mol(n),

P $\propto$ ${Z \over V}$

$\Rightarrow$ ${{{P_A}} \over {{P_B}}} = {{{Z_A}} \over {{Z_A}}} \times {{{V_B}} \over {{V_A}}}$

= $\left( {{3 \over 1}} \right) \times \left( {{1 \over 2}} \right)$ = ${3 \over 2}$

$\therefore$ 2PA = 3PB
4

### JEE Main 2019 (Online) 12th January Evening Slot

An open vessel at 27oC is heated until two fifth of the air (assumed as an ideal gas) it has escaped from the vessel assuming that the volume of the vessel remains constant, the temperature at which the vessel has been heated is -
A
750 K
B
500oC
C
750oC
D
500 K

## Explanation

We know,

PV = nRT

As the vessel is open, so the pressure and volume is constant.

$\therefore$   n1 R T1 = n2 R T2

$\Rightarrow$   n1 T1 = n2 T1

${2 \over 5}$th of the air escape from the vessel. So the remaining air is ${3 \over 5}$ of the total air.

$\therefore$   n $\times$ 300 = ${3 \over 5}$n $\times$ T2

$\Rightarrow$   T2 = 500 K